Is $x$ irreducible in $(\mathbb Z/{12\mathbb Z})[x]$?

Question

I have a couple of doubts regarding irreducibility of $x$ over $Z/{12Z}$.

First is w.r.t definition of irreducibility :

When we talk of a polynomial $p(x)$ being irreducible in $Z_{n}$, we basically mean to say that $p(x)=0$ has no solutions in $Z_{n}$ right?

If so then $x$ is not irreducible in $Z_{n}$.

I arrived at this result by showing that $A=(x)$ the ideal generated by $x$ is not maximal by proving that $Z_{12}[x]/A$ is not a field.

My attempt is as follows:

The elements of $Z_{12}[x]/(x)$ are of the form $c+A$ where $c \in Z_{12}$. But since $Z_{12}$ is not a field, $c^{-1}$ might not exist. Hence $Z_{12}[x]/(x)$ is not a field thus $x$ is not irreducible over $Z_{12}$.

Could someone check whether my attempts are right?

EDIT:

I know that a polynomial $p\in R[x]$ is said to be irreducible if it cannot be factored into the product of polynomials $f,g \in R[x]$ such that one of them is a constant.

"irreducible" is not the same as "no roots". Linear polynomials, over $\mathbb Z$ say, have roots but are irreducible. $(x^2+1)(x^2+2)$ has no roots but it is not irreducible. — lulu, Nov 24 '19 at 15:39
irreducible means having no factors right? Doesn't it essentially translate into not having roots? — Ananthakrishna, Nov 24 '19 at 15:40
@Ananthakrishna They are related, sure. They are not, however, the same thing. — Arthur, Nov 24 '19 at 15:42
@J.W.Tanner , okay! but is the approach taken by me in showing that $x$ is not irreducible is correct ? — Ananthakrishna, Nov 24 '19 at 16:02
@Ananthakrishna your approach is good and you can generalize it to any non ID ring $R[x]/(x) = R$ so $(x)$ is not maximal ideal so $x$ is reducible. — IrbidMath, Nov 24 '19 at 16:16
You should state which definition of "irreducible" and "associate" you are using since they bifurcate into inequivalent notions in rings with zero divisors, e.g. see here. — Bill Dubuque, Nov 24 '19 at 18:55

IrbidMath · Accepted Answer · 2019-11-27T22:54:52.953

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Irreducible means the following: we say that $r\in R $ is irreducible if $r = ab $ then either $a$ or $b$ is a unit. So in $\mathbb Z_{12}$, for example, $2$ is irreducible but $3$ is not since $3 = 3(-3) \mod 12 $ in general we have $p\mid n $ is reducible iff $p^2 \nmid n$.

So let's return to our ring $\mathbb{Z}_{12}[x]$ and consider $x=(3x+4 )(4x+3)= 12x^2 + 25x + 12 \bmod 12 $ hence $x$ is reducible (if we proved the factors are non units) as you said. Now we need to show that $3x+4$ and $4x+3$ are not units in $\mathbb{Z}_{12} [x]$.

You can show that $3x+4$ is a not a unit by contradiction.

edited Nov 27 '19 at 22:54

answered Nov 24 '19 at 15:53

IrbidMath

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Beware $ $ There is no standard definition of "irreducible", "associate" etc. in rings with zero divisors, e.g. see here. – Bill Dubuque Nov 24 '19 at 18:53
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Note that the above factorization of $x$ has already been mentioned here many times in the past, e.g. five years ago here, which includes citations on expositions on factorization theory in non-domains. – Bill Dubuque Aug 26 '22 at 20:38

Is $x$ irreducible in $(\mathbb Z/{12\mathbb Z})[x]$?

1 Answers1